Question

CsBr crystallises in a BCC lattice. The unit cell of length is 436.6pm given that the atomic mass of Cs is 133 amu and of Br is 80 amu and the density of CsBr is

Answer 1

Answer:

The density of CsBr is $8.50 g/cm^3$.

Explanation:

Number of atom in unit cell = Z = 4

Density of CsBr = ?

Edge length of cubic unit cell= a

Atomic mass of Pt(M) = 195.08 g/mol

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell

M = Molar mass of compound

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

We have:

M =133 g/mol + 80 g/mol = 213 g/mol

$a = 436.6 pm = 4.366\times 10^{-8} cm$

1 pm = $10^{-10} cm$

Z = 2

On substituting all the given values , we will get the value of 'a'.

$\rho=\frac{2\times 213 g/mol}{6.022\times 10^{23} mol^{-1}\times (4.366\times 10^{-8} cm)^{3}}$

$\rho = 8.50 g/cm^3$

The density of CsBr is $8.50 g/cm^3$.

Answer 2

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